# Show that a process is a local martingale

We have the following setting

We have a 2-dimensional Brownian motion $$(X,Y)$$, and we define the process $$M_t$$ as $$M_t=e^{X_t}\cos(Y_t)$$

The problem is to show that the process $$M_t$$ is a local martingale, and that we have that $$\langle M\rangle_t=\int^t_0e^{X_s}ds$$

I know that $$M_t$$ would be a local martingale if it is a continuous, adapted process and if there exists a sequence of stopping times $$(\tau_n)_{n\geq0}$$, such that $$\tau_n\uparrow\infty$$ almost surely, and that $$(M_{\tau_n\wedge t}-M_0)_{t\geq0}$$ is a uniformly integrable martingale for all $$n$$.

However, I am not aware of any straight forward way to proving that something is a local martingale, and I do not see an easy approach to proving it directly from the definition. Are there ''standard'' routes to follow when trying to prove that a process is/isn't a local martingale?

Moreover, when trying to show that the quadratic variation of $$M$$ can be written in such a way, I cannot figure out why the cosine term would drop out.

Any help is appreciated!

If we apply Ito's lemma to the function $$f(x,y) = e^x \cos(y)$$ you'll find that $$dM_t = M_t dX_t - e^{X_t} \sin(Y_t) dY_t$$ since $$\langle X \rangle_t = \langle Y \rangle_t = t$$ and $$\langle X,Y \rangle_t = 0$$. Since the stochastic integral against a continuous local martingale is again a local martingale, this shows that $$M_t$$ is a continuous local martingale. We also deduce from this that $$d \langle M \rangle_t = M_t^2 d \langle X \rangle_t + e^{2X_t} \sin(Y_t)^2 d\langle X \rangle_t = e^{2X_t} (\cos^2(Y_t) + \sin^2(Y_t))dt = e^{2X_t} dt$$ (where we have again used that $$\langle X,Y \rangle_t = 0$$) and so $$\langle M \rangle_t = \int_0^t e^{2X_s} ds$$
• Sure. Written out fully $dM_t = M_t dX_t - e^{X_t} \sin(Y_t) dY_t$ means $M_t - M_0 = \int_0^t M_s dX_s - \int_0^t e^{X_s}\sin(Y_s) dY_s$ and so $\langle M \rangle_t = \bigg \langle \int_0^t M_s dX_s - \int_0^t e^{X_s}\sin(Y_s) dY_s \bigg \rangle_t$. Now use bilinearity and symmetry of the quadratic variation since $\langle \int_0^t F_s dN_s, K_s \rangle_t = \int_0^t F_s d \langle N,K \rangle_s$ for continuous local martingale $N,K$ and an adapted process $F$ such that $\int F dN$ makes sense. Note that we don't get any cross terms between $X$ and $Y$ since $\langle X,Y \rangle =0$ – Rhys Steele Apr 1 '19 at 15:22
• I've just seen the edit to your comment. It looks like you didn't know about the behaviour of the Ito integral with respect to the quadratic variation. This is an important property to keep in mind. In fact $\int_0^t F_s dN_s$ is the unique process such that for all continuous local martingales $K_t$ we have $\langle \int_0^t F_s dN_s, K_s \rangle_t = \int_0^t F_s d \langle N,K \rangle_s$ (and one can even use this observation to give an alternative construction of the Ito integral). – Rhys Steele Apr 1 '19 at 15:25
• Oh shoot, the $t$ in my comment should be a $\cdot$, I just copy-pasted that expression and forgot to change it. The thing inside the covariation should always be a process and not a fixed random variable; that $\cdot$ just means you consider the integral as a process and not the integral at any fixed time. You're completely right to point this out. Sorry for the source of confusion. – Rhys Steele Apr 1 '19 at 15:40
• It's nothing to do with that. Try applying Ito's lemma to both of them and then use the resulting formula to compute the covariation. Recall that $\langle X,Y \rangle = 0$ and that if $V_t$ is a finite variation process then $\langle M,V \rangle = 0$ for any semimartingale $M$. – Rhys Steele Apr 3 '19 at 18:08